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C
10
H
12
O
2
in CDCl
3
Herausforderung des Monats
November 2022
In welcher Frucht stellt diese Verbindung die
Hauptgeruchskomponente dar?
Bisher gibt es zwar keine bekannte Korrelation
zwischen spektroskopischen Daten und Geruch,
aber die fertige Struktur könnte die Frage
beantworten …
13
C{
1
H} NMR-Spektrum
gemessen bei 125.83{500.36} MHz
1
H NMR-Spektrum
gemessen bei 500.13 MHz
Peak-Markierungen
< 10 - [ppm]
> 1000 - [Hz]
1
H/
1
H COSY
gemessen bei 500.13 MHz
1
H/
13
C HSQC
gemessen bei 500.13/125.77 MHz
1
H/
13
C HMBC
gemessen bei 500.13/125.77 MHz
Aus der bekannten Summenformel – beispielsweise mittels eines hochauflösenden
Massenspektrums – ergeben sich 5 Doppelbindungsäquivalente.
Erste Schritte
Doppelbindungsäquivalente,
Integration
C
10
H
12
O
2
in CDCl
3
Halten wir den Wert auf einer – hier virtuellen – Haftnotiz fest..
Die Integration ist einfach? Alle Signalgruppen sind gut getrennt. Die
Standardprozedur sollte gut funktionieren.
5 DBE
Integrale [willkürliche Einheiten (arbitrary units)]
1.00 4.91 4.86 2.37 5.12 5.12 7.38 3.68
Summe der Integrale [arbitrary units] 34.44
Proportionalitätskoeffizient [a.u. / H] 2.87
Anzahl der Protonen 0.35 1.71 1.69 0.83 1.78 1.78 2.57 1.28
C
10
H
12
O
2
in CDCl
3
Das Ergebnis ist natürlich Unfug. Auch bei großzügigster Rundung kommt man auf
keine vernünftige Anzahl an Protonen.
Aber was stimmt nicht?
5 DBE
Erste Schritte
Doppelbindungsäquivalente,
Integration
Unser Lösungsmittel zeigt trotz Deuterierung auch NMR-Signale.
C
10
H
12
O
2
in CDCl
3
Vom Lösungsmittel stammen
- das Signal des nichtdeuterierten Anteils (CHCl
3
) bei etwa 7.26 ppm und
- ein Wassersignal (entweder H
2
O oder HOD) bei etwa 1.6 … 1.7 ppm.
Die Qualität des hier verwendeten Lösungsmittels scheint nicht sehr gut zu sein. Man sieht starke
Signale der beiden oben erwähnten Komponenten. In der Routine-NMR-Spektroskopie kann das
durchaus vorkommen.
Beim zweiten Versuch der Integration ignorieren wir diese beiden Signale.
CHCl
3
H
2
O /
HOD
5 DBE
Erste Schritte
Doppelbindungsäquivalente,
Integration
Integrale [arbitrary units] 4.91 4.86 2.37 5.12 5.12 7.38
Summe der Integrale [arbitrary units] 29.76
Proportionalitätskoeffizient [a.u. / H] 2.48
Anzahl der Protonen 1.98 1.96 0.96 2.06 2.06 2.98
C
10
H
12
O
2
in CDCl
3
Die Qualität der Integration könnte noch etwas besser sein, dennoch lassen sich die 12
Protonen mühelos den verbleibenden 6 Signalgruppen zuordnen.
5 DBE
2 21 322
Erste Schritte
Doppelbindungsäquivalente,
Integration
5 DBE
2 21 322
1376.59
1384.47
1369.65
Genauere Werte der chemischen Verschiebung der vier Multipletts sowie
einige Kopplungskonstanten erhält man aus den vergrößerten
Ausschnitten des eindimensionalen Protonenspektrums.
Hie ein Beispiel für das höchstfeldige Multiplett.
=
1384.47 𝐻𝑧 + 1369.65 𝐻𝑧
2 ∗ 500.13 𝑀𝐻𝑧
= 𝟐. 𝟕𝟓 𝒑𝒑𝒎
𝐽 ≈
1384.47 𝐻𝑧 − 1369.65 𝐻𝑧
2
≈ 𝟕. 𝟒 𝑯𝒛
2.85
7.07
6.77
2.75
≈9Hz
≈7Hz
≈7Hz
Erste Schritte
Doppelbindungsäquivalente,
Integration
(Das Signal bei 2.16 ppm ist nicht
ganz einfach als Multiplett zu
erkennen. Einstweilen gelte es als
Singulett.)
2 21 322
2.85
7.07
6.77
2.75
≈9Hz
≈7Hz
≈7Hz
5 DBE
H
Als nächstes benötigen wir die drei Signalgruppen mit Signalen kleiner
als 3 ppm und deren Integrale. Das sind sechs Werte, die wir wieder auf
Haftznotizen festhalten (chemische Verschiebungen: [ppm], Integrale
[Anzahl an Protonen]).
Momentan benötigen wir das eindimensionale Protonenspektrum nicht
mehr. Heben wir es für eine mögliche spätere Verwendung auf.
2.852.752.16
223
Um unsere ersten Bausteine zu finden, benötigen wir noch einige Signale
aus dem eindimensionalen Kohlenstoffspektrum.
Erste Schritte
Doppelbindungsäquivalente,
Integration
1
H
13
C
5 DBE
2.852.752.16
223
45.5028.91 30.20
Hier benötigen wir die drei Signale mit
den kleinsten chemischen
Verschiebungen.
Im DEPT sieht man, dass die Signale bei
45.50 ppm und 28.91 ppm zu CH
2
-
Gruppen gehören, während das Signal
bei 30.20 ppm einer Methylgruppe
zuzuordnen ist.
Auch ohne einen Blick auf das HSQC
passt dies perfekt auf die Integrale der
drei Protonensignale.
Erste Schritte
Kohlenstoffsignale
(CH, CH
3
positiv, CH
2
negativ)
HSQC
1
H
Bausteine
C
5 DBE
2.852.752.16
223
45.5028.91 30.20
Jetzt verfügen wir über eine Reihe von
informationen, um die ersten Bausteine
zu suchen.
Mit dem HSQC gelingt dies perfekt.
Übersetzung in Arbeit
Letzter Bearbeitungsstand: 30. 10. 2021
Bausteine
5 DBE
2.852.752.16
223
45.5028.91 30.20
The HSQC provides us with structural fragments. To
get their chemical shifts we need the collected data
from the one-dimensional spectra.
1
H
13
C
28.91
30.20
45.50
2.16
2.752.85
3
2
2
HSQC
Building blocks
The three cross peaks in the upper right area of the
HSQC are clearly due to CH
n
fragments with sp
3
hybridized carbon.
1
H
13
C
28.91
30.20
45.50
2.752.85
3
2
2
Let’s extract the three parts of the final molecule one
after the other.
H
3
C
30.20
2.16
H
3
C
30.20
2.16
2.16
5 DBE
HSQC
Building blocks
The three cross peaks in the upper right area of the
HSQC are clearly due to CH
n
fragments with sp
3
hybridized carbon.
1
H
13
C
28.91
30.20
45.50
2.752.85
3
2
2
Let’s extract the three parts of the final molecule one
after the other.
C
H
H
28.91
2.85
C
H
H
28.91
2.85
H
3
C
30.20
2.16
2.16
5 DBE
HSQC
Building blocks
The three cross peaks in the upper right area of the
HSQC are clearly due to CH
n
fragments with sp
3
hybridized carbon.
1
H
13
C
28.91
30.20
45.50
2.752.85
3
2
2
Let’s extract the three parts of the final molecule one
after the other.
C
H
H
45.50
2.75
C
H
H
28.91
2.85
H
3
C
30.20
2.16
C
H
H
45.50
2.75
2.16
5 DBE
HSQC
Building blocks
Connectivity
We now have the first three fragments to build our
molecule. But is there any connectivity between
these fragments? The HSQC provides us the building
blocks only.
1
H
13
C
28.91
30.20
45.50
2.752.85
3
2
2
One possibility to get connectivities is the HMBC.
Let’s try.
C
H
H
28.91
2.85
H
3
C
30.20
2.16
C
H
H
45.50
2.75
2.16
5 DBE
HSQC
HMBC
HSQC
C
H
H
28.91
2.85
H
3
C
30.20
2.16
C
H
H
45.50
2.75
1
H
13
C
HSQC
5 DBE
2.85
2.75 2.16
2.85
2.75
2.16
This subset of the HMBC is sufficient and makes the
evaluation of this HMBC rich of cross peaks easier.
Very often it is a real challenge to find the really
helpful HMBC cross peaks out of the huge number of
available signals.
We already know three of the proton signals.
HMBC
Building blocks
Connectivity
C
H
H
28.91
2.85
H
3
C
30.20
2.16
C
H
H
45.50
2.75
1
H
13
C
HSQC
5 DBE
2.85
2.75 2.16
In the lower right corner of the HMBC there are three
cross peaks showing a correlation of all three proton
signals with the same carbon atom at 208.84 ppm
(the exact value is extracted from the one
dimensional carbon spectrum).
208.84
This, of course, is a carbonyl group.
C
O
208.84
HMBC
Building blocks
Connectivity
C
H
H
28.91
2.85
H
3
C
30.20
2.16
C
H
H
45.50
2.75
1
H
13
C
HSQC
5 DBE
2.85
2.75 2.16
The carbon atom of the carbonyl group is two or three
bonds away from each of the three proton groups.
208.84
C
O
208.84
It is easy to create a substructure to explain the
neighbourhoods 208.84 ppm/2.16 ppm and 208.84
ppm/2.85 ppm (via two bonds).
HMBC
Building blocks
Connectivity
C
H
H
28.91
2.85
H
3
C
30.20
2.16
C
H
H
45.50
2.75
1
H
13
C
HSQC
5 DBE
2.85
2.75 2.16
There are no more open (free) bond at the carbonyl
group. But if we join both methylene groups, the
distance between the carbon at 208.84 ppm and the
proton at 2.75 ppm is three bonds, which is totally
fine to explain an HMBC peak.
208.84
C
O
208.84
C
H
H
C
OH
3
C
208.84
30.20
28.91
2.16
2.85
HMBC
Building blocks
Connectivity
C
H
H
45.50
2.75
1
H
13
C
HSQC
5 DBE
2.85
2.75 2.16
There are no more open (free) bond at the carbonyl
group. But if we join both methylene groups, the
distance between the carbon at 208.84 ppm and the
proton at 2.75 ppm is three bonds, which is totally
fine to explain an HMBC peak.
208.84
C
H
H
C
OH
3
C
208.84
30.20
28.91
2.16
2.85
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
45.50
28.91
2.16
2.85
2.75
HMBC
Building blocks
Connectivity
1
H
13
C
HSQC
5 DBE
2.85
2.75 2.16
The carbonyl group contains one double bond
equivalent. Now there are four remaining double
bond equivalents to be assigned.
208.84
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
45.50
28.91
2.16
2.85
2.75
4 DBE
The connectivity of two methylene groups should be
visible in the COSY.
Let’s check.
HMBC
Building blocks
Connectivity
COSY
1
H
13
C
HSQC
2.85
2.75 2.16
Let’s label the pseudo projections with the three
signal groups of interest.
208.84
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
45.50
28.91
2.16
2.85
2.75
4 DBE
HMBC
HMBC
Building blocks
Connectivity
HMBC
1
H
13
C
HSQC
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
45.50
28.91
2.16
2.85
2.75
2.16
2.85
2.75
2.16
2.85
2.75
2.16
2.85
2.75
2.16
2.85
2.75
Although the chemical shifts of the two methylene
groups are similar, the connectivity between both
groups is clearly visible in the COSY.
4 DBE
Building blocks
Connectivity
HMBC
1
H
13
C
HSQC
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
45.50
28.91
2.16
2.85
2.75
2.16
2.85
2.75
2.16
2.85
2.75
But there is another really unexpected correlation
visible.
4 DBE
Building blocks
Connectivity
HMBC
1
H
13
C
HSQC
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
45.50
28.91
2.16
2.85
2.75
2.16
2.85
2.75
2.16
2.85
2.75
Using our temporary structure, this cross peak would
be the result of a coupling pathway across five single
bonds. Such coupling pathways are really very, very
rare.
Even across four single bonds coupling is not visible
very often, but apparently we have such a case in our
compound. If we swap both methylene groups, the
coupling distance reduces to four bonds.
4 DBE
Building blocks
Connectivity
HMBC
1
H
13
C
HSQC
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
2.16
2.16
2.85
2.75
2.16
2.85
2.75
Now we have four single bonds between the
coupling nuclei.
45.50
28.91
2.85
2.75
But if there is a coupling, our methyl signal should be
a triplet?
4 DBE
Building blocks
Connectivity
HMBC
13
C
HSQC
2.16
2.85
2.75
2.16
2.85
2.75
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
C
C
H
H
H
C
OH
3
C
H
200.84
30.20
28.91
45.50
2.16
2.75
2.85
COSY
1
H
Indeed! The methyl protons show a triplet with a
four bond coupling constant of 0.45 Hz.
4 DBE
Building blocks
Connectivity
2 21 322
7.07
6.77
≈9Hz
HMBC
13
C
HSQC
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
COSY
4 DBE
Indeed! The methyl protons show a triplet with a
four bond coupling constant of 0.45 Hz.
0.45 Hz
5.166.777.07
122
≈9Hz
Let us use the proton spectrum to extract the
data for the three signal groups with chemical
shifts larger than 5 ppm as done before with the
signals below 3 ppm. Additional we should
remember the coupling constant of roughly 9 Hz.
Building blocks
Connectivity
2 21 322
7.07
6.77
≈9Hz
HMBC
13
C
HSQC
Building blocks
One more substituent
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
COSY
4 DBE
For the sake of clarity let us neglect the tiny coupling
constant. We no longer need this value to get our
structure
0.45 Hz
5.166.777.07
122
≈9Hz
The signal at 5.15 ppm is certainly worth a closer
examination. A rather limited amount of NMR
signals appear in this area.
To do so, let's take another look at the HSQC.
2 21 322
7.07
6.77
≈9Hz
HMBC
13
C
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
COSY
4 DBE
5.156.777.07
122
≈9Hz
H
HSQC
As said, we are interested in the peak at 5.16 ppm.
Building blocks
One more substituent
HMBC
13
C
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
COSY
4 DBE
5.166.777.07
122
≈9Hz
1
H
5.16
1
There is no cross peak at all with the proton signal at
5.16 ppm.
Because the molecular formula contains only H, C
and O this means, this signal belongs to an OH group.
O
H
5.16
HSQC
Building blocks
One more substituent
HSQC
HMBC
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
5.16
1
There is no cross peak at all with the proton signal at
5.16 ppm.
Because the molecular formula contains only H, C
and O this means, this signal belongs to an OH group.
O
H
5.16
13
C
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
There are four signals in the one dimensional carbon spectrum, which are
not part of the already found substructures.
Let‘s create four more sticky notes with the chemical shifts of these signals.
HSQC
Building blocks
One more substituent
(CH, CH
3
positive, CH
2
negative)
HMBC
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
O
H
5.16
HSQC
153.98 132.95 129.43 115.34
(Numbers in italics refer to the
low-intensity signals, numbers in
bold refer to the intense signals.)
Now it‘s time for an intermediate inventory.
Building blocks
One more substituent
(CH, CH
3
positive, CH
2
negative)
HMBC
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
O
H
5.16
HSQC
153.98 132.95 129.43 115.34
C
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
Building blocks
One more substituent
HMBC
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
O
H
5.16
HSQC
153.98 132.95 129.43 115.34
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
13
C
Our molecular formula is C
10
H
12
O
2
.
We found two fragments. Together their molecular formula is C
4
H
8
O
2
.
What else do we know?
• We have only two proton signal groups for four protons. The
chemical shift is typical for protons bound to sp
2
hybridized
carbon atoms.
• We have four carbon signals but six carbon atoms.
• The missing fragment contains 4 double bond equivalents.
Clearly there is some kind of symmetry.
Maybe there are exotic solutions to fulfill all these conditions.
But one possible structure stands out. What about benzene,
substituted in the 1 and 4 positions? Two possible substituents have
already been identified.
missing: C
6
H
4
Building blocks
The missing link
HMBC
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
O
H
5.16
HSQC
153.98 132.95 129.43 115.34
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
13
C
C
C
C
C
C
C
H
HH
H
132.95
153.98
129.43
115.34
7.07
6.77
And the signal assignments fell from sky?
They didn‘t, of course. And even more, they might be wrong.
But first let‘s finalize our molecule. The explanation for the
assignment will be given afterwards.
Maybe there are exotic solutions to fulfill all these conditions.
But one possible structure stands out. What about benzene,
substituted in the 1 and 4 positions? Two possible substituents have
already been identified.
Building blocks
The missing link
HMBC
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
O
H
5.16
153.98 132.95 129.43 115.34
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
13
C
C
C
C
C
C
C
H
HH
H
132.95
153.98
129.43
115.34
7.07
6.77
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
129.43
115.34
2.16
2.75
2.85
7.07
6.77
5.16
Now that we know the final structure with some signal
assignment fallen from the sky, let's take another look at the
HSQC.
HSQC
Building blocks
The missing link
HMBC
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
153.98 132.95 129.43 115.34
13
C
HSQC
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
129.43
115.34
2.16
2.75
2.85
7.07
6.77
5.16
For both pseudo projections we
need the chemical shifts
corresponding to these cross
peaks
Signal assignment
HMBC
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
153.98 132.95 129.43 115.34
13
C
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
129.43
115.34
2.16
2.75
2.85
7.07
6.77
5.16
For both pseudo projections we
need the chemical shifts
corresponding to these cross
peaks
115.34
129.43
7.07 6.77
6.777.07
HSQC
Signal assignment
HMBC
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
153.98 132.95
13
C
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
129.43
115.34
2.16
2.75
2.85
7.07
6.77
5.16
Now we can confirm a part of
the assignment that fell from
the sky before, using the two
HSQC cross peaks. The two CH
groups can of course be
interchanged.
115.34
129.43
7.07 6.77
The assignment of the two
quaternary carbon atoms could
of course also be reversed.
HSQC
Signal assignment
HMBC
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
153.98 132.95
13
C
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
129.43
115.34
2.16
2.75
2.85
7.07
6.77
5.16
Now we can confirm the
assignment that fell from the
sky before, using the two HSQC
cross peaks. The two CH groups
can of course be interchanged.
115.34
129.43
7.07 6.77
The assignment of the two
quaternary carbon atoms could
of course also be reversed.
HSQC
HSQC
Signal assignment
HMBC
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
13
C
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
129.43
115.34
2.16
2.75
2.85
7.07
6.77
5.16
The protons of the aromatic ring form a non-trivial AA'XX'
spin system. Nevertheless, there is a dominant vicinal
coupling constant.
The protons with chemical shifts of 6.77 ppm and 7.07
ppm are in the Z-position to each other. The coupling
constant should be about 8 ... 10 Hz.
We have extracted a value of approximately this size from
the multiplets of the two protons.
And, of course, there is a cross peak in the COSY between
the two proton signals mentioned above.
There is a last argument supporting our six
membered aromatic ring.
HSQC
Signal assignment
A good possibility to check the assignments made
for the atoms of the six membered aromatic ring is
the HMBC.
COSY
1
H
13
C
HSQC
HMBC
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
129.43
115.34
2.16
2.75
2.85
7.07
6.77
5.16
Signal assignment
COSY
1
H
13
C
HSQC
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
129.43
115.34
2.16
2.75
2.85
7.07
6.77
5.16
2.85 2.75
5.16
132.95
153.98
2.75
2.85
5.16
132.95
153.98
To check the correct assigment of
the quaternary carbon atoms we
need only a few peaks.
HMBC
Signal assignment
COSY
1
H
13
C
HSQC
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
129.43
115.34
2.16
2.75
2.85
7.07
6.77
5.16
2.85 2.75
5.16
132.95
153.98
Everything is fine. The
cross peaks come from
two and three bond
correlations.
This wouldn‘t be true with
the opposite assignment
of the quaternary carbon
atoms.
But what about the carbon signals at
129.43 ppm and 115.34 ppm?
Let‘s try.
HMBC
Signal assignment
Building blocks
COSY
1
H
13
C
HSQC
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
129.43
115.34
2.16
2.75
2.85
7.07
6.77
5.16
2.85 2.75
5.16
The pseudo projection for the
protons is fine, but we need other
values for the carbon pseudo
projection.
129.43
115.34
129.43
115.34
In both cases the coupling pathway is
across four bonds.
That‘s not really impossible, but maybe
there is another possibility.
Let‘s change the CH assignment inside
the six membered ring.
HMBC
COSY
1
H
13
C
HSQC
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
2.16
2.75
2.85
5.16
2.85 2.75
5.16
129.43
115.34
129.43
115.34
7.07
6.77
HMBC
Signal assignment
COSY
1
H
13
C
HSQC
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
115.34
129.43
2.16
2.75
2.85
6.77
7.07
5.16
2.85 2.75
5.16
129.43
115.34
Now our cross peaks come from
coupling pathways across three bonds,
which is much more probable.
Fortunately, there is a last proof
supporting this assignment.
HMBC
Signal assignment
1
H
13
C
HSQC
HMBC
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
115.34
129.43
2.16
2.75
2.85
6.77
7.07
5.16
COSY
We didn‘t investigate one cross peak of
the COSY. Let's catch up.
HMBC
Signal assignment
Building blocks
1
H
13
C
HSQC
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
115.34
129.43
2.16
2.75
2.85
6.77
7.07
5.16
HMBC
2.85
7.07
2.85
7.07
We have to label two signals groups only.
And now there is a weak cross peak via four bonds. This
would be five bonds with the opposite assignment of
the CH groups.
Final solution
1
H
13
C
HSQC
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
115.34
129.43
2.16
2.75
2.85
6.77
7.07
5.16
HMBC
COSY
The primary question was about the smell of
the compound.
Without a dictionary, of course, you can
not answer the question.
Raspberries
